# A thermometer is taken from a room where the temperature is 20^o C to the outdoors, where the...

## Question:

A thermometer is taken from a room where the temperature is {eq}20^{\circ} C {/eq} to the outdoors, where the temperature is {eq}5^{\circ} C {/eq} . After one minute the thermometer reads {eq}12^{\circ} C {/eq} .

(a) What will the reading on the thermometer be after one more minute?

(b) When will the thermometer read {eq}6^{\circ} C {/eq} ?

## Newton's Law of Cooling:

Newto's laew of cooling states that " the rate of change of temperature of an object is proportional to the difference of the temperature of the object and the temperature of its surroundings. "

According to the law, the temperature of a body at time 't' is given by:

{eq}\displaystyle { T_t = T_s + (T_0 - T_s) e^{-kt} } {/eq}

where,

{eq}T_s {/eq} is the temperature of the surroundings

{eq}T_0 {/eq} is the initial temperature.

{eq}k {/eq} is the constant.

Given:

• Initial temperature, {eq}T_0 = 20^{\circ} C {/eq}
• Surrounding temperature, {eq}T_s = 5^{\circ} C {/eq}
• After 1 minute that is 60 seconds the temperature is, {eq}T_{60} = 12^{\circ} C {/eq}

According to the Newton's law of cooling the temperature at time 't' is given by:

{eq}\displaystyle { T_t = T_s + (T_0 - T_s) e^{-kt} \\ \Rightarrow 12 = 5 + (20 - 5) e^{-60k} \\ \Rightarrow \frac{7}{15} = e^{-60k} \\ \Rightarrow ln \frac{7}{15} = \ln e^{-60k} \\ \Rightarrow -0.762 = -60k \\ \Rightarrow k = 0.0127 } {/eq}

(a)

After 1 more minute that is 60 second, the temperature would be,

{eq}\displaystyle { T_{120} = 5 + (20 - 5) e^{-0.0127*120} \\ = 5 + 15*0.2178 \\ = 8.26712^{\circ} C } {/eq}

(b)

Time at which the thermometer reading is {eq}\ 6^{\circ} C {/eq} is :

{eq}\displaystyle { 6 = 5 + (20 - 5) e^{-0.0127*t} \\ \Rightarrow 1 = 15e^{-0.0127*t} \\ \Rightarrow \frac{1}{15} = e^{-0.0127*t} \\ \Rightarrow \ln \frac{1}{15} = \ln e^{-0.0127*t} \\ \Rightarrow -2.708 = -0.0127*t \\ \Rightarrow t = 213.228 \approx 214 \ seconds } {/eq}